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Question

A 3 kg ball is dropped from the roof of a building 176.4 m high. While the ball is falling to earth, a horizontal wind exerts a constant force of 12 N on the ball. How long does it take to hit the floor? And how far from the building does the ball hit the ground?
Answer: 6 seconds, and 72 meters

Answer 1

First you must realize that the ball will hit the ground in the same time regardless of any horizontal force. So calculating the time to impact totally ignores any horizontal influence. On the other hand, WHERE it hits the ground has nothing to do with any vertical force such as gravity, except for the simple fact of determining exactly how long the ball is "in flight", which then enables you to calculate how far afield it will be when it hits. The previous answers very adequatly describe how this is done.

Answer 2

First, look at the vertical motion, and use the equation of motion that deals with uniform acceleration, like a falling ball:
d = vt + 1/2 at^2

where d is distance = 176.4 m
v = initial speed = 0 (assuming it is dropped, not thrown)
t= time = ?
a = acceleration = 9.8 m/s^2 in free fall on Earth

so
176.4 = 0 + 1/2(9.8)t^2
352.8 = 9.8t^2
36 = t^2
t = 6 seconds

So the ball falls for 6 seconds. How far does it move horizontally?

Use the formula F = ma
12 N = 3kg*a
a = 4 m/s^2

and use the first formula again
d = vt +1/2at^2
d = 0 + 1/2(4)(6)^2
d= 72 meters

Answer 3

in a physics question, you can easily simplify the problem by looking at only one axis at a time (the x- or y- axis) meaning horizontal or vertical

let's start with vertical:
the weight of the ball does not matter because everything falls at the same rate (acceleration of gravity)
you know the change in y (distance) is 176.4 m
you know the initial velocity is 0 m/s (it is dropped, not thrown)
you know the acceleration of gravity is 9.8 m/s^2
and you need to find the time

the equation you will use is
(change in y) = (intital velocity)(time) + (1/2)(acceleration)(time)^2
so plug in the numbers and find time, the easiest way to rearrange would be
time = [-(inital velocity) +/- [(initialvelocity)^2 + 4(- change in y)(1/2)(acceleration)]^(1/2)] / [2(1/2)(acceleration)]

through the equation
x = [-b +/- [(b^2) + 4ac]^(1/2)] / 2a

now the horizontal:
for the horizontal, you know the only force exerted on the ball is the wind
you will use the equation
sigma F = mA
i will put E instead of sigma to simplify things (sigma simply means the sum of)
so
EF = mA
12 = 3A
A = 4 m/s^2
(change in x) = (initial velocity)(time) + (1/2)(acceleration)(time)^2
simply plug in the numbers you have (inital velocity is zero and time is the same as in the vertical direction) to find the change in x, or the distance in the horizontal plane

Answer 4

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Answer 5

First you need to know the acceleration due to gravity which is 9.8 or 10 for easier equations. Using equations which you should really learn for physics problems, you get the time to be 6 seconds. Then if there is 12 N of force horizontally, then multiply it by the time in the air which is 6 seconds. Then you get 72 meters.

Answer 6

An object falling is accelerated by gravity at 9.81m/s^2. The formula for distance under constant acceleration is:
s = v1 + 1/2at^2
s = 176.4
v1 = initial velocity = 0
a = acceleration = 9.81
t = time
Solve for t. I'll let you do the work.

For horizontal distance:
f=ma
f = 12N
m = 3kg
a = acceleration

Substitute into:
s = v1 + 1/2at^2
v1 = 0
a = acceleration calculated above
t = time from above

Do the math, if I haven't forgotten anything since high school, you should get the correct answer.

EDIT: PoisonMushroom is treating the 12N as a constant speed. Your question indicates it is a force causing acceleration.